Activity 34: - Mathematical sciences
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Transcript Activity 34: - Mathematical sciences
ACTIVITY 34
Review (Sections 3.6+3.7+4.1+4.2+4.3)
Problems 3 and 5:
Sketch the graph of the function by transforming an appropriate
function of the form y = xn. Indicate all x- and y-intercepts on
each graph.
P( x) ( x 2)
3
y - intercept
y xx3 23
y
y x 2
3
x0
P0 (0 2)3 8
x - intercept
y0
0 ( x 2)3
0 ( x 2)3
0 ( x 2)
2,0
x 2
0,8
P( x) 2 x 8
4
No x – intercepts
y – intercepts is (0,8)
y = x4
y = 2x4
y = 2x4 + 8
Problems 11, 13, and 15:
Match the polynomial function with the below graphs
P( x) x( x2 4) R( x) x5 5x3 4 x
T ( x) x 4 2 x3
T ( x) x 4 2 x3
R( x) x5 5x3 4 x
P( x) x( x2 4)
Problems 21:
Sketch the graph of the polynomial function. Make sure your
graph shows all intercepts and exhibits the proper end behavior.
P( x) x 3x 23x 2
x - intercepts/ zero's
x3
The leading term is 3x3
x 2
2
x
3
y - intercept
x0
P(0) 0 30 230 2
32 2
12
Problem 31:
Factor the polynomial P(x) = −x3 + x2 + 12x and use the
factored form to find the zeros. Then sketch the graph.
P( x) x3 x 2 12x
x x 2 x 12
xx 4x 3
Zeros
x0
x0
x 4 0
x4
x 3 0
x 3
Problem 3:
Let P(x) = x4 − x3 + 4x + 2 and Q(x) = x2 + 3.
(a) Divide P(x) by Q(x).
(b) Express P(x) in the form P(x) = D(x) · Q(x) + R(x).
x2 – x – 3
x2 +
x4 − x3 + 4x + 2
3
Q (x )
–xx4 –+ 3x
3x22
P( x ) Q ( x ) D( x ) R( x )
x4 x3 4 x 2 x2 x 3x2 3 7 x 11
− x3 – 3x2 + 4x + 2
+– x33
+
– 3x
3x
– 3x2 + 7x + 2
– 3x22
+3x
+
–9
7x + 11
R (x )
Problem 9:
Find the quotient and remainder using long division for the expression
x3 6 x 3
x2 2 x 2
P( x ) Q ( x ) D( x ) R( x )
x +2
x2 – 2x+2
P( x ) Q ( x ) D( x ) R( x )
D( x )
D( x )
x3 + 6x + 3
–xx3 +
– 2x
2x22 +– 2x
Q ( x ) D( x ) R( x )
R( x )
Q( x)
D( x )
D( x )
D( x )
x3 6 x 3
8x 1
x 2 2
2
x 2x 2
x 2x 2
Q (x )
2x2 + 4x + 3
– 2x2 –+ 4x
4x +– 4
8x – 1
R (x )
Problem 21:
Find the quotient and remainder using synthetic division for the
expression
3 1
x3 8x 2
x3
1
Q( x) x 3x 1
R( x ) 1
2
0
8
2
3
9
3
3
1
1
Q (x )
R (x )
Problem 35:
Use synthetic division and the Remainder Theorem to evaluate
P(c) if P(x) = 5x4 + 30x3 − 40x2 + 36x + 14 and c = −7.
7
5
5
30 40
14
35 35
35 497
5 5
71 483
Q (x )
P( c ) R( c )
P( 7) 483
36
R (x )
Problem 43:
Use the Factor Theorem to show that x − 1 is a factor of
P(x) = x3 − 3x2 + 3x − 1.
1
1 3
1
3 1
1
2
1
2
1
0
Showing that x = 1 is a zero.
Therefore, (x – 1) is a factor
Problem 53:
Find a polynomial of degree 3 that has zeros 1, −2, and 3,
and in which the coefficient of x2 is 3.
a x 1 x 2 x 3
ax 2 x x 2x 3
2
ax x 2x 3
ax3 3x2 x2 3x 2 x 6
2
ax3 2 x 2 5x 6
ax 2ax 5ax 6a
3
2
2a 3
3
a
2
3
x 1x 2x 3
2
Problem 3:
List all possible rational zeros given by the Rational Zeros Theorem (but don’t
check to see which actually are zeros).
R( x) 2 x5 3x3 4 x2 8
divisors of 8 1, 2, 4, 8
p
1, 2
divisors of 2
q
possible rational zeros:
1, 1 , 2, 4, 8
2
Problems 13 and 23:
Find all rational zeros of the polynomial.
P( x) x3 3x 2
divisors of 2 1, 2
p
1
divisors of 1
q
possible rational zeros:
x
3
1, 2
3 2 1 1
3x 2 x x 2 x (1)
x12 x 12 x 12
1
1
1
02
1
2 4
1
0
1
3 2
1
1 2
2
0
x3 3x 2 x2 x 2 x 1
x
2
x 2 x 2x 1
x3 3x 2 x 2x 1x 1
Consequently, the zero’s
are x = 2 and x = – 1
P( x) x 4 6 x3 7 x2 6 x 8
divisors of 8 1, 2, 4, 8
p
divisors of 1
q
1
possible rational zeros: 1,
2, 4, 8
P( x) x 4 6 x3 7 x2 6 x 8
1
1
1
possible rational zeros: 1,
6
7
6
8
1
7
14
8
7
14
8
0
3
2
x 4 6 x 3 7 x 2 6 x 8 x 7 x 14x 8 x 1
So we need only factor
x
3
7 x2 14x 8
2, 4, 8
possible rational zeros: 1, 2, 4, 8
possible rational zeros: 1,
2, 4, 8
x
3
7 x2 14x 8
1 1
1
possible rational zeros: 1,
2, 4, 8
7
14 8
1 6 8
6
8
0
x 6 x 8x (1)x 1
x 6 x 8x 1x 1
x 6x 8 x 4x 1
3
2
x 4 6 x 3 7 x 2 6 x 8 x 7 x 14x 8 x 1
2
2
So we need only factor
2
x4 6x3 7 x2 6x 8 x 4x 2x 1x 1
Consequently, the roots are x = -4, x = -2, x = -1, and x = 1